1. TIF 21101
APPLIED MATH 1
(MATEMATIKA TERAPAN 1)
Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng
Week 4
Relation and Function I
2. Relation and FunctionRelation and Function
Overview
Obviously, we do not realize that there many connections
are happened in our circumtances. For examples, day and
night happens because of earth rotation, all students in
math are also connected to other subjects and so on.
Strictly speaking, something happens because of other
Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng
Strictly speaking, something happens because of other
subject called “reason”.
Relations can be used to solve problems such as
determining which pairs of cities are linked by airline flights
in a network, finding a viable order for the different phases
of a complicated project, or producing a useful way to store
information in computer databases.
For couple weeks later, you all will be introduced this
“connection” in mathematic’s view. And we shall learn to
“map” or “transform” the “connection”.
3. Relation and FunctionRelation and Function
Objectives
Cartesian Product
Relation
Invers Relation
Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng
Invers Relation
Pictoral Repesentation of Relation
Composition of Relation
Relation Properties
4. Relation and FunctionRelation and Function
Cartesian Product
Consider two sets A and B. The set of all ordered
pairs (a, b) where a∈A and b∈B is called the
Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng
pairs (a, b) where a∈A and b∈B is called the
product, or Cartesian product, of A and B.
The short designation of this product is A x B,
which is read “A cross B”.
5. Relation and FunctionRelation and Function
Ex.
Let A={1, 2} and B={a, b, c}.
Then
AxB {(1,a},(1,b),(1,c),(2,a),(2,b),(2,c)}
BxA {(a, 1), (a,2), (b, 1), (b,2), (c,1),(c,2)}
AxA {(1, 1), (1,2), (2,1), (2,2)}
From the example above we can conclude, that,
First,
Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng
First,
A x B ≠≠≠≠ B x A
The Cartesian product deals with ordered pairs, so naturally the order in
which the sets are considered is important.
Second, using n(s) for the number of elements in a set S, we have
n(A x B) = n(A) . n(B) = 2 x 3 = 6
Therefore, there will be 26 = 64 relation from A to B
So…..what is relation?????
6. Relation and FunctionRelation and Function
Relation
Relation is just a subset of the Cartesian product
of the sets.
Definition.
Let A and B be sets. A binary relation or, simply,
Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng
Let A and B be sets. A binary relation or, simply,
relation from A to B is a subset of A x B.
In other words, a binary relation from A to B is a
set R of ordered pairs where the first element
(domain) of each ordered pair comes from A and
the second element (codomain or range) comes
from B.
7. Relation and FunctionRelation and Function
We use the notation a R b to denote that (a, b) ∈ R
and a R b to denote that (a, b) ∉ R.
Moreover, when (a, b) belongs to R, a is said to be
related to b by R.
/
Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng
Assume C= {1,2,3} and D ={x,y,z} and let R {(1,y), (1,z),
(3,y)}. Put the R or R for the followings:
1…X 1…Y 1…Z
2…X 2…Y 2…Z
3…X 3…Y 3…Z
/
8. Relation and FunctionRelation and Function
Invers Relation
The invers relation of set is defined as the opposite
mapping of relation itself.
Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng
Let R be any relation from a set A to a set B. The
inverse of R, denoted by R-1, is the relation from B
to A which consists of those ordered pairs which,
when reversed, belong to R; that is,
R-1= {(b,a): (a,b) ∈ R}
9. Relation and FunctionRelation and Function
Ex.
Let R = {(1,y), (1,z), (3,y)} from A = {1,2,3} to
B = {x,y,z}, then
Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng
R-1 = {(y, 1), (z, 1), (y,3)}
10. Relation and FunctionRelation and Function
Pictoral Repesentation of Relation
Arrow Diagram
Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng
11. Relation and FunctionRelation and Function
Table Representation
Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng
12. Relation and FunctionRelation and Function
Matrice Representation
Suppose R is the relation from A to B, where
A={ a1,a2,a3,…,am} and B={ b1,b2,b3,…,bn}.
The relation can be describe in matrice M=[mij] as
Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng
The relation can be describe in matrice M=[mij] as
folow:
13. Relation and FunctionRelation and Function
Ex.
a1 = 2
a2 = 3
a3 = 4
Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng
a3 = 4
b1 = 2
b2 = 4
b3 = 8
b4 = 9
b5 = 15
14. Relation and FunctionRelation and Function
Directed Graph
First we write down the elements of the set, and
then we drawn an arrow from each element x to
each element y whenever x is related to y.
Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng
each element y whenever x is related to y.
The point is, directed graph does not show the
relation between one set to the other. It just shows
the relation among the element inside the set.
Ex. R is relation on the set A = {1,2,3,4}
R = {(1,2), (2,2), (2,4), (3,2), (3,4), (4,1), (4,3)}
16. Relation and FunctionRelation and Function
Prac.
Show the relation from
the directed graph
Bandung
Jakarta Surabaya
Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng
Medan
Makassar
Kupang
17. Relation and FunctionRelation and Function
Composition of Relation
Suppose A, B, and C be sets, and let R be a
relation from A to B and let S be a relation
from B to C. R ⊆ A x B and S ⊆ B x C.
Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng
from B to C. R ⊆ A x B and S ⊆ B x C.
Then R and S give rise to a relation from A
to C, which is denoted by RoS and defined
as
18. Relation and FunctionRelation and Function
Ex.
Assume A= {1,2,3,4}, B ={a,b,c,d}, C ={x,y,z}
and let R= {(1,a), (2,d), (3,a) (3,b), (3,d)} and
S ={(b,x), (b,z), (c,y), (d,z)} . Show the
relation a(RoS)c!
Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng
relation a(RoS)c!
19. Relation and FunctionRelation and Function
From the picture we can observe that there is an arrow
from 2 to d which is followed by an arrow from d to z. We
can view these two arrows as a “path” which “connects” the
element 2 ∈ A to the element z ∈ C. Thus,
2(R o S)z since 2Rd and dSz
Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng
Similarly there is a path from 3 to x and a path from 3 to z.
Hence,
3(R o S)x and 3(R o S)z
No other element of A is connected to an element of C.
Therefore, the composition of relations R o S gives
RoS= {(2,z), (3,x), (3,z)}
20. Relation and FunctionRelation and Function
Soal :
R = {(1, 2), (1, 6), (2, 4), (3, 4), (3, 6), (3, 8)}
S = {(2, u), (4, s), (4, t), (6, t), (8, u)}
Gambarkan grafiknya dan tentukan R o S
Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng
Gambarkan grafiknya dan tentukan R o S
21. Relation and FunctionRelation and Function
Matematika Terapan 12014/2015 M. Ilyas Hadikusuma, M.Eng
R o S = {(1, u), (1, t), (2, s), (2, t), (3, s), (3, t), (3, u) }